Answer:
a. -6.99 × 10³⁰ C/m³ b. The layer of air is negatively charged.
Explanation:
With E₁ = electric field at 500 m above the ground = 160 N/C (it is negative since it is directed downwards).
Also, with E₂ = electric field at 800 m above the ground = 120 N/C (it is negative since it is directed downwards).
The total flux, Ψ = ∫E.dA = E₁dAcosθ + E₂dAcosθ
For the 800 m surface E is parallel to dA, that is = 0° and For the 500 m surface E is anti-parallel to dA, that is = 180°
Ψ = ∫E₁dAcos180° + ∫E₂dAcos0°
= -∫E₁dA + ∫E₂dA
= -E₁∫dA + E₂∫dA
= -E₁4πR² + E₂4πR²
= (-E₁ + E₂)4πR² where R = radius of earth = 6.4 × 10⁶ m
= (-160 N/C + 120 N/C)4π(6.4 × 10⁶ m)²
= - 40 N/C)4π(6.4 × 10⁶ m)²
= -20588.74 × 10¹² C
= -2.058874 × 10¹⁶ C
≅ -2.06 × 10¹⁶ Nm²/C.
The since charge, Q = Ψ/ε₀, the total charge through the area is thus
Q = Ψ/ε₀
= -2.06 × 10¹⁶ Nm²/C ÷ 8.854 × 10⁻¹² F/m
= -0.233 × 10²⁸ C/m²
= -2.33 × 10²⁸ C/m².
So, the charge in the volume = charge net charge of surface × width of volume. So the charge in the volume Q' = QΔh = Q(h₂ - h₁) where h₁ = 500 m and h₂ = 800 m
Q' = Q(h₂ - h₁)
= -2.33 × 10²⁸ C/m²(800 m - 500 m)
= -2.33 × 10²⁸ C/m²(300 m)
= -699 × 10²⁸ C/m³
= -6.99 × 10³⁰ C/m³
b. Since Q' = -6.99 × 10³⁰ C/m³, the layer of air is negatively charged.
